Diffusion near a wall

[ronojoy]

@vavachan, lets shift focus on diffusion near a wall, just for a while. What we want is that the particle maintains a roughly steady z-distance from the wall, but is free to move in the x-y plane otherwise. Therefore, can you run simulations with a potential that is harmonic only in the z-coordinate, that is there is a z-component to the force on the particle, of the form

F(z) = -k(z - z0) 

and no other forces. Solve the Langevin equations for this system and compute the mean square displacement

C(\tau) = < |x(t + \tau) - x(t) |^2 >

and plot it as a function of \tau. The slope at long times is the diffusion coefficient. If everything is ok, this should agree with the Einstein result and diffusion should slow down as you approach the wall.

[vavachan]

[vavachan]

Comparison of numerical and analytical results for diffusion parallel to the wall

There is small variation from the analytic result of slope evaluated with gamma(z0), but that decreases with more averaging and steeped gaussian.

slope corresponding z0= 10.0 , .002529, 2_KB_T/(gamma(z0))= .002518 z0= .8 ,0.00174 2_KB_T/(gamma(z0))= .001828 z0= .4 ,0.001548 2_KB_T/(gamma(z0))= .001556 z0= .3 ,0.00128 2_KB_T/(gamma(z0))= .00144

[vavachan]

Comparison of analytical formula with friction evaluated at mean height with numerical results.

[vavachan]

Yesterday I made a mistake telling you that the radius of the particle is .1 unit, In the simulations I was using 1 unit. So all distance scales were already normalised. I ran simulation with z0 at .5 and range of z approx .5

The variation gamma_per(z0) (normalized) is

and mean square displacement for x ,

[vavachan]

There is not much variation for X , because gamma parallel doesn't vary drastically as gamma perpendicular.

but gamma parallel doesn't vary too much unless very close to the wall

both axes are normalized

[vavachan]

I have done the simulation with gavity.

m=1 g=9.8 Kb=1 Temp=20 a=1 eta =.1

This is the distribution along z. It agrees with the analytic expression.

This is the mean square displacement with hydrodynamic interaction plotted alongside MSD with constant gamma. (I have plotted data from different simulations with same parameters)

slope in the absence of hydrodynamic interaction is 20.77 which corresponds to gamma of 1.92 which agrees to the theoretical value

Slope in the presence of the hydrodynamic interaction is around 15.8 which corresponds to gamma of 2.53

[vavachan]

[ronojoy]

@vavachan, please have a look at section 5 (pg. 58) and figures 7 and 8 of Chandrasekhar's review. These give the dynamics of P( z t | z0 0) as it approaches sedimentation equilibrium. You should compute similar z height histograms from your simulation data.

[vavachan]

This is the evolution of distribution along z . It resembles the graphs given in the chandrshekhar's review. I could not get the prob_dist for later times because simulation will take a long time to finish, but I think the trend is evident, Since I know that the final distribution will be exponential. If I could do more averaging then it will be much clearer.

The 'Histogram_X.dat' X corresponds to X_1000_dt time. with my dt=.001, m=1,gamma_0()=1.88

[vavachan]

[vavachan]

http://arxiv.org/pdf/physics/9912055.pdf

[vavachan]

https://www1.maths.leeds.ac.uk/~grant/BLL.pdf

Kevin burrage's paper , I have implemented Manella's leap frog method, pg 261 in the pdf.

On Wed, Jun 24, 2015 at 12:57 AM, Ronojoy Adhikari <notifications@github.com

wrote:

@vavachan https://github.com/vavachan, please have a look at section 5 (pg. 58) and figures 7 and 8 of Chandrasekhar's review. These give the dynamics of P( z t | z0 0) as it approaches sedimentation equilibrium. You should compute similar z height histograms from your simulation data.

— Reply to this email directly or view it on GitHub https://github.com/vavachan/brownian_motion/issues/5#issuecomment-114616824 .

[vavachan]

[vavachan]

The green line is <D(z)> vs mass and red line is the slope obtained from simulations, x axis is the mass

[vavachan]

http://ac.els-cdn.com/S0375960198008238/1-s2.0-S0375960198008238-main.pdf?_tid=c7504ecc-2624-11e5-aa52-00000aab0f26&acdnat=1436437691_c88bb88764d37e00281b25ea5a124cdb

[vavachan]

[ronojoy]

@vavachan, here is a somewhat more detailed suggestion on how to go forward with the calculation.

1. Consider only the z-part of motion, which is the pair

\dot v_z = A(z)v_z + B(z)dW_z
\dot z = v_z

Adiabatically eliminate the momentum and get the Fokker-Planck equation for position alone, using Wilemski's method or that of Aguirre and Murphy.

1. Attempt to solve this Fokker-Planck equation by reducing it to a pure diffusion equation, using a generalization of the transformation that Chandrasekhar uses in his Eq. 428. Think about what should take the place of the exponential in that equation - I expect it to be the quadrature solution of the deterministic part of the Langevin equation.
2. Once you have the solution, return to the complete equations for x and z, and examine if the Fokker-Planck equation can be made to factorize. Physically, how much you go up or down doesn't depend on where you are along the plane, but how much you move sideways does depend on how far up or down you are from the plane. There must be a mathematical statement of this - you have to find it.
3. Once this is done, you can combine the results from 2 and 3 and work out the exact expression for the parallel diffusivity.

Let me know if you have any questions. Also, it might be a good idea to start assembling things into a report.